C4 Differentiation - Implicit differentiation · C4 Differentiation - Implicit differentiation...
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1. A curve C has equation
2x + y2 = 2xy
Find the exact value of xy
dd at the point on C with coordinates (3, 2).
(Total 7 marks)
2. The curve C has the equation
cos2x + cos3y = 1, 6
0,44
πππ≤≤≤≤− yx
(a) Find xy
dd in terms of x and y.
(3)
The point P lies on C where x = 6π .
(b) Find the value of y at P. (3)
(c) Find the equation of the tangent to C at P, giving your answer in the form ax + by + cπ = 0, where a, b and c are integers.
(3) (Total 9 marks)
3. The curve C has the equation ye–2x = 2x + y2.
(a) Find xy
dd in terms of x and y.
(5)
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The point P on C has coordinates (0, 1).
(b) Find the equation of the normal to C at P, giving your answer in the form ax + by + c = 0, where a, b and c are integers.
(4) (Total 9 marks)
4. A curve C has the equation y2 – 3y = x3 + 8.
(a) Find xy
dd in terms of x and y.
(4)
(b) Hence find the gradient of C at the point where y = 3. (3)
(Total 7 marks)
5. A curve has equation 3x2 – y2 + xy = 4. The points P and Q lie on the curve. The gradient of the
tangent to the curve is 38 at P and at Q.
(a) Use implicit differentiation to show that y – 2x = 0 at P and at Q. (6)
(b) Find the coordinates of P and Q. (3)
(Total 9 marks)
6. A curve is described by the equation
x3 – 4y2 = 12xy.
(a) Find the coordinates of the two points on the curve where x = –8. (3)
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(b) Find the gradient of the curve at each of these points. (6)
(Total 9 marks)
7. A set of curves is given by the equation sin x + cos y = 0.5.
(a) Use implicit differentiation to find an expression for xy
dd
.
(2)
For –π < x < π and –π < y < π,
(b) find the coordinates of the points where xy
dd
= 0.
(5) (Total 7 marks)
8. A curve C is described by the equation
3x2 – 2y2 + 2x – 3y + 5 = 0.
Find an equation of the normal to C at the point (0, 1), giving your answer in the form ax + by + c = 0, where a, b and c are integers.
(Total 7 marks)
9. A curve C is described by the equation
3x2 + 4y2 – 2x + 6xy – 5 = 0.
Find an equation of the tangent to C at the point (1, –2), giving your answer in the form ax + by + c = 0, where a, b and c are integers.
(Total 7 marks)
Edexcel Internal Review 3
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10. The value £V of a car t years after the 1st January 2001 is given by the formula
V = 10 000 × (1.5)–t.
(a) Find the value of the car on 1st January 2005. (2)
(b) Find the value of tV
dd
when t = 4.
(3)
(c) Explain what the answer to part (b) represents. (1)
(Total 6 marks)
11. A curve has equation
x2 + 2xy – 3y2 + 16 = 0.
Find the coordinates of the points on the curve where xy
dd
= 0.
(Total 7 marks)
12. The curve C has equation 5x2 + 2xy – 3y2 + 3 = 0. The point P on the curve C has coordinates (1, 2).
(a) Find the gradient of the curve at P. (5)
(b) Find the equation of the normal to the curve C at P, in the form y = ax + b, where a and b are constants.
(3) (Total 8 marks)
Edexcel Internal Review 4
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= 0.
(Total 7 marks)
13. A curve has equation
x3 − 2xy − 4x + y3 − 51 = 0.
Find an equation of the normal to the curve at the point (4, 3), giving your answer in the form ax + by + c = 0, where a, b and c are integers.
(Total 8 marks)
Edexcel Internal Review 5
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1. ( )d 2 ln 2.2d
x x
x= B1
d dln 2.2 2 2 2d d
x y yy y xx x
+ = + M1 A1 = A1
Substituting ( )3, 2
d d8ln 2 4 4 6d d
y yx x
+ = + M1
d 4 ln 2 2d
yx= − Accept exact equivalents M1 A1 7
[7]
2. (a) –2 sin 2x – 3 sin 3yxy
dd =0 M1 A1
yx
xy
3sin32sin2–
dd
= Accepty
x3sin3–
2sin2 ,
yx
3sin32sin2– A1 3
(b) At ,6π
=x 13cos6
2cos =+
yπ M1
213cos =y A1
93
3 ππ=⇒= yy awrt 0.349 A1 3
(c) At
9,
6ππ ,
( )( ) 3
2sin3sin2
3sin32sin2
–dd
3
3
9
6 −=−==π
π
π
π
xy M1
=
6–
32–
9– ππ xy M1
Leading to 6x + 9y–2π = 0 A1 3 [9]
3. (a) e–2x xyyy
xy x
dd22e2–
dd 2– += A1 correct RHS *M1 A1
Edexcel Internal Review 6
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xxx yx
yx
2–2–2– e2–dde)e(
dd
= B1
(e–2x – 2y) xyxy 2–e22
dd
+= *M1
y
yxy
x
x
2–ee22
dd
2–
–2+= A1 5
(b) At P, 4–2–ºeºe22
dd
=+
=xy M1
Using mm′ = –1
41'=m M1
)0–(411– xy = M1
x – 4y + 4 = 0 or any integer multiple A1 4
Alternative for (a) differentiating implicitly with respect to y.
e–2x – 2ye–2x yyx
yx 2
dd2
dd
+= A1 correct RHS *M1 A1
yxyy
yxxx
dde2–e)e(
dd 2–2–2– = B1
(2 + 2ye–2x) =yx
dd e–2x – 2y *M1
x
x
yy
yx
2–
2–
e222–e
dd
+=
y
yxy
x
x
2–ee22
dd
2–
2–+= A1 5
[9]
4. (a) C: y2 – 3y = x3 + 8 Differentiates implicitly to include either
23dd3–
dd2 x
xy
xyy = ±ky .)
ddIgnore(.
dd3or
dd
=±
xy
xy
xy M1
Correct equation. A1
A correct (condoning sign error) attempt to
(2y – 3) 23dd x
xy= combine or factorise their ‘ ,.
dd3–
dd2'
xy
xyy M1
Can be implied.
Edexcel Internal Review 7
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3–2
3dd 2
yx
xy=
3–23 2
yx A1 oe 4
(b) y = 3 ⇒ 9 – 3(3) = x3 + 8 Substitutes y = 3 into C. M1
x3 = – 8 ⇒ x = – 2 A1
rking.correct wo from4dd
=xy
(–2, 3) ⇒ 4dd
3–6)4(3
dd
=⇒=xy
xy Also can be ft using
their ‘x’ value and y = 3 in the A1ft 3
correct part (a) of 3–2
3dd 2
yx
xy=
(b) final A1 . Note if the candidate inserts their x value and y = 3 into
3–23
dd 2
yx
xy= , then an answer of =
xy
dd their x2, may indicate
a correct follow through. [7]
5. (a) 3x2 – y2 + xy = 4 (eqn *)
38
26
38
dd
26
ddor
26
dd
0dd
dd26
=−−−
⇒=
−+
=
−−−
=
=
++−
yxyx
xy
xyyx
xy
yxyx
xy
xyxy
xyyx
not necessarily required.
giving –18x – 3y = 8x – 16y giving 13y = 26x Hence, y = 2x ⇒ y – 2x = 0
Differentiates implicitly to include either
)dd (Ignore .
ddor
dd
=±
xy
xyx
xyky M1
Correct application )( of product rule B1
(3x2 – y2) →
−
xyyx
dd26 and (4 → 0) A1
Substituting 38
dd
=xy into their equation. M1*
Attempt to combine either terms in x or terms in y together
Edexcel Internal Review 8
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to give either ax or by. dM1*
simplifying to give y – 2x = 0 AG A1 cso 6
(b) At P & Q, y = 2x. Substituting into eqn * gives 3x2 – (2x)2 + x(2x) = 4 Simplifying gives, x2 = 4 ⇒ x = ±2 y = 2x ⇒ y = ±4 Hence coordinates are (2, 4) and (–2, –4)
Attempt replacing y by 2x in at least one of the y terms in eqn * M1
Either x = 2 or x = –2 A1
Both (2, 4) and (–2, –4) A1 3 [9]
6. (a) x3 – 4y2 = 12xy (eqn *) x = –8 ⇒ –512 – 4y2 = 12(–8)y –512 – 4y2= –96y
Substitutes x = –8 (at least once) into * to obtain a three term quadratic in y. Condone the loss of = 0. M1
4y2 – 96y + 512 = 0 y2 – 24y + 128 = 0
(y – 16)(y – 8) = 0
2)128(457624 −±
=y
y = 16 or y = 8.
An attempt to solve the quadratic in y by either factorising or by the formula or by completing the square. dM1
Both y = 16 and y = 8. or (–8, 8) and (–8, 16). A1 3
(b) 3x2 – 8y
+=
xyxy
xy
dd1212;
dd
+−
=yxyx
xy
812123
dd 2
@ (–8, 8), 332
96)8(8)8(12)8(12)64(3
dd
−=−
=+−−
=xy
,
Edexcel Internal Review 9
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@ (–8, 16), 0320
)16(8)8(12)16(12)64(3
dd
==+−−
=xy
.
Differentiates implicitly to include either xyx
xyky
dd12or
dd
± .
Ignore ...dd
=xy
M1
Correct LHS equation; A1;
Correct application of product rule (B1)
not necessarily required.
Substitutes x = –8 and at least one of their y-values to attempt
to find any one of xy
dd
. dM1
One gradient found. A1
Both gradients of –3 and 0 correctly found. A1 cso 6
Aliter Way 2
+=− x
yxyy
xyx 12
dd12;8
dd3 2
+−
=yxyx
xy
812123
dd 2
@ (–8, 8), 332
96)8(8)8(12)8(12)64(3
dd
−=−
=+−−
=xy
,
@ (–8, 16), 0320
)16(8)8(12)16(12)64(3
dd
==+−−
=xy
.
Differentiates implicitly to include either yxy
yxkx
dd12or
dd2± .
Ignore ...dd
=yx M1
Correct LHS equation; A1;
Correct application of product rule (B1)
not necessarily required.
Substitutes x = –8 and at least one of their y-values to attempt
to find any one of yx
xy
ddor
dd
. dM1
One gradient found. A1
Both gradients of –3 and 0 correctly found. A1 cso 6
Edexcel Internal Review 10
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Aliter Way 3
x3 – 4y2 = 12xy (eqn *) 4y2 + 12xy – x3 = 0
21
32
32
32
32
)9(21
23
89412
81614412
8))(4(414412
xxxy
xxxy
xxxy
xxxy
+±−=
+±−=
+±−=
−−±−=
21
32
2
221
32
)9(4
31823
dd
)318(;)9(21
21
23
dd
xx
xxxy
xxxxxy
+
+±−=
++
±−=
−
@ x = –8 21
))512()64(9(4
)64(3)8(1823
dd
−+
+−−=
xy
.0,323
23
dd
3248
23
)64(448
23
−=±−=∴
±−=±−=
xy
A credible attempt to make y the subject and an attempt to
differentiate either 21
32 )9(21or
23 xxx +− . M1
))g(()9(23
dd 2
132 xxxk
xy −
+±−= A1
)318(;)9(21
21
23
dd 22
132 xxxx
xy
++
±−=
− A1
Substitutes x = –8 find any one of xy
dd
. dM1
One gradient correctly found. A1 Both gradients of –3 and 0 correctly found. A1 6
[9]
7 (a) sin x + cos y = 0.5 (eqn *)
Edexcel Internal Review 11
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cos x – sin y xy
dd
= 0 (eqn #)
yx
xy
sincos
dd
=
Differentiates implicitly to include ± sin y xy
dd
. (Ignore (xy
dd
=).) M1
yx
sincos A1 cso 2
(b) 0cos0sincos0
dd
=⇒=⇒= xyx
xy
giving 2π
−=x or 2π
=x
When x = 5.0cos2
sin,2
=+
−− yππ
When x = 5.0cos2
sin,2
=+
yππ
⇒ cos y = 1.5 ⇒ y has no solutions
⇒ cos y = – 0.5 ⇒ y = 3
2or 3
2 ππ−
In specified range (x, y) =
−
32,
2 and
32,
2ππππ
Candidate realises that they need to solve ‘their numerator’ = 0
…or candidate sets xy
dd
= 0 in their (eqn #) and attempts to solve
the resulting equation. M1ft
both 2
,2
ππ−=x or x = ±90° or awrt x = ± 1.57 required here A1
Substitutes either their 2
or 2
ππ−== xx into eqn * M1
Only one of 3
2-or 3
2 ππ=y or 120°
or –120° or awrt –2.09 or awrt 2.09 A1
Only exact coordinates of
−
32,
2 and
32,
2ππππ A1
Do not award this mark if candidate states other coordinates inside the required range. A1 5
[7]
Edexcel Internal Review 12
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Edexcel Internal Review 13
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8. dydx = 0
dd32
dd46 =−+−
xy
xyyx
Differentiates implicitly to include either xyor
xyky
dd3
dd
±± .
(ignore
=
xy
dd .) M1
Correct equation. A1
++
=3426
dd
yx
xy
not necessarily required.
At (0, 1), 72
3420
dd
=++
=xy
Substituting x = 0 & y = 1 into an equation involving ;dd
xy dM1
to give 72 or
72
−− A1 cso
Hence m(N) 27
−= or 72
1− A1ftoe.
Uses m(T) to ‘correctly’ find m(N). Can be ft from “their tangent gradient”.
Either N: y – 1 = –27 (x – 0)
or: N: 127 +−= xy
)0(1 −=− xmy + with ‘their tangent or normal gradient’; or uses y = mx + 1 with ‘their tangent or normal gradient’ ; M1;
N: 7x + 2y − 2 = 0 Correct equation in the form ‘ax + by + c = 0’, A1 oe where a, b and c are integers. cso
[7]
Edexcel Internal Review 14
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Beware: 72
dd
=xy does not necessarily imply the award of all the first four marks in this
question.
So please ensure that you check candidates’ initial differentiation before awarding the first A1 mark.
Beware: The final accuracy mark is for completely correct solutions. If a candidate flukes the final line then they must be awarded A0.
Beware: A candidate finding an m(T) = 0 can obtain A1ft for m(N) = ∞, but obtains M0 if they write y − 1 = ∞( x − 0). If they write, however, N: x = 0, then can score M1.
Beware: A candidate finding an m(T) = ∞ can obtain A1ft for m(N) = 0, and also obtains M1 if they write y − 1 = 0(x − 0)or y = 1.
Beware: The final cso refers to the whole question.
Aliter Way 2
dxdy = 03
dd24
dd6 =−+−
yxy
yxx
Differentiates implicitly to include either yx
yxkx
dd2or
dd
±±
(ignore
=
yx
dd .) M1
Correct equation. A1
++
=2634
dd
xy
yx
not necessarily required.
At (0, 1), 27
2034
dd
=++
=yx
Substituting x = 0 & y = 1 into an equation involving yx
dd ; dM1
to give 27 A1 cso
Edexcel Internal Review 15
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Hence m(N) = 27 or
72
1− A1ftoe.
Uses m(T) or yx
dd to ‘correctly’ find m(N).
Can be ft using “−1 .yx
dd ”.
Either N: y − 1 = )0(27 −− x M1
or N: y = 127 +− x
y − 1 = m(x − 0) with ‘their tangent, yx
dd or normal gradient’;
or uses y = mx + 1 with ‘their tangent, yx
dd or normal
gradient’;
N: 7x + 2y − 2 = 0 Correct equation in the form ‘ax + by + c = 0’, A1oe where a, b and c are integers. cso
[7]
Aliter Way 3
2y2 + 3y − 3x2 − 2x – 5 = 0
( ) 25
23
169
43 22
++=−+ xy x
43
1649
23 )2
−++= xy x Differentiates using the chain rule; M1
( ) )13(21
dd 2
12
1649
23 +++=
−xx
xy x
Correct expression for xy
dd ; A1 oe
Edexcel Internal Review 16
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At (0, 1),
72
74
21
1649
21
dd 2
1
=
=
=
−
xy
Substituting x = 0 into an equation involving xy
dd ; dM1
to give 72 or 7
2−− A1 cso
Hence m(N) = 27
− A1ft
Uses m(T) to ‘correctly’ find m(N). Can be ft from “their tangent gradient”.
Either N: y − 1 )0(27 −− x
or N: y 172 +−= x M1
y − 1 = m(x − 0) with ‘their tangent or normal gradient’; or uses y = mx + 1 with ‘their tangent or normal gradient’
N: 7x + 2y − 2 = 0 A1 oe Correct equation in the form ‘ ax + by + c = 0’, where a, b and c are integers.
[7]
9. Differentiates
to obtain: 6x + 8yxy
dd – 2,
.................................. + (6x xy
dd + 6y) = 0 +(B1)
+−−
=yx
yxxy
86662
dd
Edexcel Internal Review 17
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Substitutes x = 1, y = –2 into expression involving xy
dd , to give
xy
dd = –
108 M1, A1
Uses line equation with numerical ‘gradient’ y – (– 2) = (their gradient) M1 (x – 1) or finds c and uses y = (their gradient) x + “c”
To give 5y + 4x + 6 = 0 (or equivalent = 0) A1ft [7]
10. (a) Substitutes t = 4 to give V, = 1975.31 or 1975.30 or 1975 or 1980 (3 s.f) M1, A1 2
(b) tV
dd = –ln1.5 × V; = –800.92 or –800.9 or –801 M1 A1; A1 3
M1 needs ln 1.5 term
(c) rate of decrease in value on 1st January 2005 B1 1 [6]
11. 2x + xyyy
xyx
dd62
dd2 −
+ = 0 M1 (A1) A1
xy
dd = 0 ⇒ x + y = 0 (or equivalent) M1
Eliminating either variable and solving for at least one value of x or y. M1 y2 – 2y2 – 3y2 + 16 = 0 or the same equation in x y = ±2 or x = ±2 A1 (2 – 2), (–2, 2) A1 7
Note: xyyx
xy
−+
=3d
d
Alternative:
3y2 – 2xy – (x2 + 16) = 0
y =6
)19216(2 2 +± xx
)19216(
8.31
31
dd
2 +±=
x
xxy M1 A1 ± A1
xy
dd = 0 ⇒
)19216(
82 +x
x = ±1 M1
64x2 = 16x2 + 192 M1 A1 x = ±2 A1 7 (2, –2), (–2, 2)
[7]
Edexcel Internal Review 18
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12. (a) 10x, +(2y + 2xxy
dd ), –6y
xy
dd = 0 M1, (B1), A1
At (1 , 2) 10 + (4 + 2xy
dd ) – 12
xy
dd = 0 M1
∴1014
dd
=xy = 1.4 or
57 or 1 5
2 A1 5
(b) The gradient of the normal is 75
− M1
Its equation is y – 2 = 75
− (x – 1) M1
(allow tangent)
y = 75
− x + 275 or y =
75
− x + 7
19 A1cao 3
[8]
13. Differentiates w.r.t. x to give
3x2, − 2xxy
dd + 2y, −4 + 3y2
xy
dd = 0 M1, B1, A1
At (4, 3)
48 − (8y′ + 6) − 4 + 27y′ = 0 M1
⇒ y′ = −1938 = −2 A1
∴Gradient of normal is 21 M1
∴y − 3 = 21 (x − 4) M1
i.e. 2y − 6 = x − 4
x – 2y + 2 = 0 A1 8 [8]
Edexcel Internal Review 19
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1. This question was also well answered and the general principles of implicit differentiation were well understood. By far the commonest source of error was in differentiating 2x; examples such as 2x, 2x lnx and x2x–1 were all regularly seen. Those who knew how to differentiate 2x
nearly
always completed the question correctly, although a few had difficulty in finding )2(dd xyx
correctly. A minority of candidates attempted the question by taking the logs of both sides of the printed equation or a rearrangement of the equation in the form 2x = 2xy – y2. Correctly done, this leads to quite a neat solution, but, more frequently, errors, such as ln(2x + y2) = ln 2x + ln y2, were seen.
It was noteworthy that a number of correct solutions were seen using partial differentiation, a topic which is not in the A level Mathematics or Further Mathematics specifications. These were, of course, awarded full marks.
2. As has been noted in earlier reports, the quality of work in the topic of implicit differentiation has improved in recent years and many candidates successfully differentiated the equation and
rearranged it to find xy
dd . Some, however, forgot to differentiate the constant. A not infrequent,
error was candidates writing xy
dd = –2sin 2x – 3sin3y
xy
dd and then incorporating the
superfluous xy
dd on the left hand side of the equation into their answer. Errors like
xy
dd (cos3y)
= – y3sin31 .
were also seen. Part (b) was very well done. A few candidates gave the answer 20° , not recognising that the question required radians. Nearly all knew how to tackle part (c) although a few, as in Q2, spoilt otherwise completely correct solutions by not giving the answer in the form specified by the question.
3. As noted above work on this topic has shown a marked improvement and the median mark scored by candidates on this question was 8 out of 9. The only errors frequently seen were in differentiating xey 2– implicitly with respect to x. A few candidates failed to read the question correctly and found the equation of the tangent instead of the normal or failed to give their answer to part (b) in the form requested.
4. A significant majority of candidates were able to score full marks on this question. In part (a), many candidates were able to differentiate implicitly and examiners noticed fewer candidates differentiating 8 incorrectly with respect to x to give 8. In part (b), many candidates were able to substitute y = 3 into C leading to the correct x-coordinate of –2. Several candidates either rearranged their C equation incorrectly to give x = 2 or had difficulty finding the cube root of –8. Some weaker candidates did not substitute y = 3 into C, but substituted y = 3 into the x
ydd
expression to give a gradient of x2.
Edexcel Internal Review 20
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5. This question was generally well done with a majority of candidates scoring at least 6 of the 9 marks available.
In part (a), implicit differentiation was well handled with most candidates appreciating the need to apply the product rule to the xy term. A few candidates failed to differentiate the constant
term and some wrote “xy
dd
= ...” before starting to differentiate the equation. After
differentiating implicitly, the majority of candidates rearranged the resulting equation to make
xy
dd
the subject before substituting xy
dd
as 38 rather than substituting
38 for
xy
dd
in their
differentiated equation. Many candidates were able to prove the result of y – 2x = 0. A
surprising number of candidates when faced with manipulating the equation 38
26
=−+
xyyx
,
separated the fraction to incorrectly form two equations 6x + y = 8 & 2y – x = 3 and then proceeded to solve these equations simultaneously.
Some candidates, who were unsuccessful in completing part (a), gave up on the whole question even though it was still possible for them to obtain full marks in part (b). Other candidates, however, did not realise that they were expected to substitute y = 2x into the equation of the curve and made no creditable progress with this part. Those candidates who used the
substitution y = 2x made fewer errors than those who used the substitution x = 2y
. The most
common errors in this part were for candidates to rewrite – y2 as either 4x2 or –2x2; or to solve the equation x2 = 4 to give only x = 2 or even x = ±4. On finding x = ±2, some candidates went onto substitute these values back into the equation of the curve, forming a quadratic equation and usually finding “extra” unwanted points rather than simply doubling their two values of x to find the corresponding two values for y. Most candidates who progressed this far were able to link their values of x and y together, usually as coordinates.
6. This question was generally well done with many candidates scoring at least seven or eight of the nine marks available.
In part (a), the majority of candidates were able to use algebra to gain all three marks available with ease. It was disappointing, however, to see a significant minority of candidates at A2 level who were unable to correctly substitute y = –8 into the given equation or solve the resulting quadratic to find the correct values for y.
In part (b), implicit differentiation was well handled, with most candidates appreciating the need to apply the product rule to the 12xy term although errors in sign occurred particularly with those candidates who had initially rearranged the given equation so that all terms were on the LHS. A few candidates made errors in rearranging their correctly differentiated equation to
make xy
dd
the subject. Also some candidates lost either one or two marks when manipulating
their correctly substituted xy
dd
expressions to find the gradients.
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C4 Differentiation - Implicit differentiation PhysicsAndMathsTutor.com
7. In part (a), the majority of candidates were able to successfully differentiate the given equation
to obtain a correct expression for xy
dd
, although there were a small proportion of candidates who
appeared to “forget” to differentiate the constant term of 0.5. Some candidates, as was similar with Q3, produced a sign error when differentiating sin xand cos y with respect to x. These
candidates then went on to produce the correct answer for xy
dd
, but lost the final accuracy mark.
A few candidates incorrectly believed that the expression yx
sincos could be simplified to give
cot x.
In part (b), the majority of candidates realised that they needed to set their numerator equal to
zero in order to solve xy
dd
= 0. Most candidates were then able to obtain at least one value for x,
usually x = 2π , although x =
2π
− was not always found. A surprising number of candidates
did not realise that they then needed to substitute their x value(s) back into the original equation in order for them to find y. Of those who did, little consideration was given to find all the
solutions in the specified range, with a majority of these candidates finding y = 3
2π , but only a
minority of candidates also finding y = 3
2π− . Therefore it was uncommon for candidates to
score full marks in this part. Some candidates also incorrectly set their denominator equal to zero to find extra coordinates inside the range. Also another small minority of candidates stated
other incorrect coordinates such as
−
32,
2ππ or
−−
32,
2ππ in addition to the two sets of
coordinates required. These candidates were penalised by losing the final accuracy mark.
8. This question was successfully completed by the majority of candidates. Whilst many demonstrated a good grasp of the idea of implicit differentiation there were a few who did not
appear to know how to differentiate implicitly. Candidates who found an expression for dxdy in
terms of x and y, before substituting in values of x= 1 and y = 1, were prone to errors in manipulation. Some candidates found the equation of the tangent and a number of candidates did not give the equation of the normal in the requested form.
9. `This question was generally well answered with most candidates showing good skills in
differentiating explicitly. Candidates who found an expression for yx
dd in terms of x and y,
before substituting in values, were more prone to errors in manipulation. Some candidates found the equation of the normal and a number of candidates did not give the equation of the tangent in the requested form. It was quite common to see such statements as
Edexcel Internal Review 22
C4 Differentiation - Implicit differentiation PhysicsAndMathsTutor.com
06dd62–
dd86
dd
=
+++= y
xyx
xyyx
yx , but often subsequent correct working indicated that this
was just poor presentation.
10. (a) Most understood the context of this problem and realised that they needed to use t = 4, although t = 0, 1 or 5 were often seen.
(b) Very few had any idea at all about how to differentiate V (many gave their answer as –t(1.5)-t-1, or had a term (1.5)-2t).
(c) The comments made in answer to the request to interpret their answer to part (b) were usually too generalised and vague. The examiners required a statement that the value of dVdt
which had been found represented the rate of change of value on 1st January 2005
11. Almost all candidates could start this question and the majority could differentiate implicitly correctly. This is an area of work which has definitely improved in recent years. Many, having
founddd
yx
, could not use it and it was disturbing to find a substantial number of students in this
relatively advanced A2 module proceeding from 03x yy x+
=−
to x + y = 3y – x. Those who did
obtain y = –x often went no further but those who did could usually obtain both correct points, although extra incorrect points were often seen.
12. This was usually well done, but differentiation of a product caused problems for a number of
candidates. Many still insisted on making dydx
the subject of their formula before substituting
values for x and y. This often led to unnecessary algebraic errors.
13. No Report available for this question.
Edexcel Internal Review 23